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HELP ME HELP ME GEOMETRY PROBLM!!!

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In triangle ABC, m<A = 90, m<B=75, and BC = \sqrt{3}. What is the area of ABC? Express your answer as a common fraction.

HELP THIS IS SO HARD HHELP MEE!!! EXplain too please!

  • HELP ME HELP ME GEOMETRY PROBLM!!! -

    You have a right-angled triangle so we can use the formula
    (1/2) base x height

    Since you want the answer as a common fraction, I will assume that you need "exact" values.

    you state that BC = \sqrt{3}
    I will assume that is meant to say √3
    so cos75 =base/√3
    base = √3cos75
    and sin75 = height/√3
    height = √3sin75

    area = (1/2)(√3sin75)(√3cos75)
    = (3/2)sin75cos75)
    (recall sin 2A = 2sinAcosA)
    = (3/2)(2sin75cos75)/2
    = (3/4) sin 150

    sin 150 = sin30 ---- one of the basic identities

    area = (1/3) sin150
    = (3/4)sin30
    = (3/4)(1/2) = 3/8

  • HELP ME HELP ME GEOMETRY PROBLM!!! -

    thanks alot

  • HELP ME HELP ME GEOMETRY PROBLM!!! -

    THANK YOU !!!!!!!

  • HELP ME HELP ME GEOMETRY PROBLM!!! -

    You don't even need trig for this. Reflect B, draw and altitude from B to B'C and solve

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